Answer:
The Symmetric Property states that for all real numbers x and y , if x = y , then y = x . Transitive Property. The Transitive Property states that for all real numbers x , y , and z , if x = y and y = z , then x = z . Substitution Property. If x = y , then x may be replaced by y in any equation or expression. I think it is.
Step-by-step explanation:
Polygon ABCD is plotted on a coordinate plane and then rotated 90 clockwise about point C to form polygon A’B’C’D Match each vertex of polygon ABCD to its coordinates.
Answer:
A' - (8,2)
B' - (5,1)
C' - (4,2)
D' - (4,5)
Step-by-step explanation:
See attachment for the missing figure.
We can see that the vertices of the polygon ABCD have coordinates A(4,6), B(5,3), C(4,2) and D(1,2)
Polygon ABCD is rotated 90° clockwise about point C to form polygon A′B′C′D′ (see attached diagram), then
A'(8,2);
B'(5,1);
C' is the same as C, thus, C'(4,2);
D'(4,5).
Please help Solving linear and quadratic equations
Answer: B.
x ≈2.5
Step-by-step explanation:
[tex]-\left(u\right)^{-1}-6=-u+10[/tex]
[tex]u=8-\sqrt{65},\:u=8+\sqrt{65}[/tex]
[tex]x=\frac{\ln \left(8+\sqrt{65}\right)}{\ln \left(3\right)}[/tex]
x=2.52...
Answer:
x=2.5
Step-by-step explanation:
The equation of a circle is x2 + y2 = 56.25. Find the radius of the circle?
Answer:
r = 7.5
Step-by-step explanation:
Circle equation: [tex](x - h)^2 + (y - k)^2 = r^2[/tex]
Since we are already give r², we simply just take the square root of 56.25, and we should get 7.5 as our final answer!
Find the Laplace transform F(s)=L{f(t)} of the function f(t)=sin2(wt), defined on the interval t≥0. F(s)=L{sin2(wt)}= help (formulas) Hint: Use a double-angle trigonometric identity. For what values of s does the Laplace transform exist? help (inequalities)
The Laplace transform of the function [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex] .
The Laplace transform exist when s > 0 .
Here, the given function is f(t) = sin²(wt) .
The Laplace transform of the the function f(t),
F(s) = f(t) = { [tex]{\frac{1}{2} \times 2sin^2(wt) }[/tex] }
F(s) = { [tex]\frac{1}{2} \times (1- cos2wt)[/tex] }
F(s) = { [tex]\frac{1}{2} - \frac{1}{2} \times cos(2wt)\\[/tex] }
F(s) = [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex]
Next,
The above Laplace transform exist if s > 0 .
Know more about Laplace transform,
https://brainly.com/question/31481915
#SPJ4
In a random sample of 7 residents of the state of Maine, the mean waste recycled per person per day was 1.4 pounds with a standard deviation of 0.23 pounds. Determine the 95% confidence interval for the mean waste recycled per person per day for the population of Maine. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.
Answer:
[tex]df=n-1=7-1=6[/tex]
The Confidence desired is 0.95 or 95%, then the significance is [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and the critical value in the t distribution with 6 degrees of freedom would be [tex]t_{\alpha/2}=2.447[/tex]
[tex]1.4-2.447\frac{0.23}{\sqrt{7}}=1.187[/tex]
[tex]1.4+2.447\frac{0.23}{\sqrt{7}}=1.613[/tex]
The 95% confidence interval for the true mean would be between (1.187;1.613)
Step-by-step explanation:
Information given
[tex]\bar X =1.4[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=0.23 represent the sample standard deviation
n=7 represent the sample size
Confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom are given by
[tex]df=n-1=7-1=6[/tex]
The Confidence desired is 0.95 or 95%, then the significance is [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and the critical value in the t distribution with 6 degrees of freedom would be [tex]t_{\alpha/2}=2.447[/tex]
Now replacing we got:
[tex]1.4-2.447\frac{0.23}{\sqrt{7}}=1.187[/tex]
[tex]1.4+2.447\frac{0.23}{\sqrt{7}}=1.613[/tex]
The 95% confidence interval for the true mean would be between (1.187;1.613)
T-Mobile sells 6 different models of cell phones and have found that they sell an equal number of each model. The probability distribution that would describe this random variable is called:
Answer:
Option A is correct.
A uniform distribution.
Step-by-step explanation:
Complete Question
T-Mobile sells 6 different models of cell phones and have found that they sell an equal number of each model. The probability distribution that would describe this random variable is called:
A) Uniform Distribution
B) Continuous Distribution
C) Poisson Distribution
D) Relative Frequency Distribution
Solution
A uniform distribution is one in which all the variables have the same probability of occurring.
It is also known as a rectangular distribution, as every portion of the sample space has an equal chance of occurring, with equal length on the probability curve, leading to a rectangular probability curve.
And for this question, 6 different models of phones sell an equal number, hence, the probability of selling each model is equal to one another, hence, this is evidently a uniform distribution.
Hope this Helps!!!
The manager of the Danvers-Hilton Resort Hotel stated that the mean guest bill for a weekend is $600 or less. A member of the hotel's accounting staff noticed that the total charges for guest bills have been increasing in recent months. The accountant will use a sample of future weekend guest bills to test the manager's claim. (a) Which form of the hypotheses should be used to test the manager's claim? H0: - Select your answer - Ha: - Select your answer - The member of the hotel's accounting staff suspects that the total charges for guest bills have Select in recent months. To test the manager’s claim, the staff member will conduct Select test of the population Select . (b) What conclusion is appropriate when H0 cannot be rejected? When H0 cannot be rejected, there Select enough evidence to conclude that the total charges for guest bills have Select in recent months. (c) What conclusion is appropriate when H0 can be rejected? When H0 can be rejected, there Select enough evidence to conclude that the total charges for guest bills have Select in recent m
Answer:
a) Null hypothesis (H0): the mean guest bill for a weekend is $600.
Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.
b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600.
c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.
Step-by-step explanation:
a) The accountant, as he wants to see if there is evidence to support the claim that the mean guest bill has increased significanty, should write the hypothesis like that:
Null hypothesis (H0): the mean guest bill for a weekend is $600.
Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.
A sample of bills of the period in study needs to be taken in order to have a representation of the actual population of bills and then perform a t-test, as the sample mean and standard deviation will be used to perform the test.
b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600. If the P-value was low but not enough, they may take another sample to perform the test again or leave it like that.
c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.
) For which values of x is f '(x) zero? (Enter your answers as a comma-separated list.) x = (No Response) For which values of x is f '(x) positive? (Enter your answer using interval notation.) (No Response) For which values of x is f '(x) negative? (Enter your answer using interval notation.) (No Response) What do these values mean? f is (No Response) when f ' > 0 and f is (No Response) when f ' < 0. (b) For which values of x is f ''(x) zero? (Enter your answers as a comma-separated list.)
Answer:
Check below, please
Step-by-step explanation:
Step-by-step explanation:
1.For which values of x is f '(x) zero? (Enter your answers as a comma-separated list.)
When the derivative of a function is equal to zero, then it occurs when we have either a local minimum or a local maximum point. So for our x-coordinates we can say
[tex]f'(x)=0\: at \:x=2, and\: x=-2[/tex]
2. For which values of x is f '(x) positive?
Whenever we have
[tex]f'(x)>0[/tex]
then function is increasing. Since if we could start tracing tangent lines over that graph, those tangent lines would point up.
[tex]f'(x)>0 \:at [-4,-2) \:and\:(2, \infty)[/tex]
3. For which values of x is f '(x) negative?
On the other hand, every time the function is decreasing its derivative would be negative. The opposite case of the previous explanation. So
[tex]f'(x) <0 \: at\: [-2,2][/tex]
4.What do these values mean?
[tex]f(x) \:is \:increasing\:when\:f'(x) >0\\\\f(x)\:is\:decreasing\:when f'(x)<0[/tex]
5.(b) For which values of x is f ''(x) zero?
In its inflection points, i.e. when the concavity of the curve changes. Since the function was not provided. There's no way to be precise, but roughly
at x=-4 and x=4
PLS HELP I GIVE BRAINLIEST
A car is travelling at 70 mph and a lorry is travelling at 50mph
calculate the speed of the car relative to the lorry
Answer:
Step-by-step explanation:
relative speed is calculated by the difference of the one with respect to the other
let car be 70 mph
and lorry be 50 mph../
relative speed of car with respect to lorry = 70-50 = 20 mph
this is ur answer
The speed of the car relative to the lorry when a car and a lorry travelling-
in the same direction : 20 mph
in the opposite direction: 120 mph
What is relative speed?"Relative speed is the speed of one body with respect to another."
Formula for relative speed: -If two bodies moving with speed m, n (m > n)
in the same direction, then the relative speed = m - n
in the opposite direction, then the relative speed = m + n
Given: - a car is travelling at 70 mph and a lorry is travelling at 50 mph.
If car and a lorry travelling in the same direction then the relative speed would be,
= 70 - 50
= 20 mph
If the car and a lorry travelling in the opposite direction then the relative speed would be,
= 70 + 50
= 120 mph
Therefore, the speed of the car relative to the lorry is 20 mph if they are travelling in the same direction and 120 mph if the car and a lorry travel in opposite direction.
Learn more about relative speed here:
https://brainly.com/question/24692222
#SPJ2
Any help would be great
Answer:
2/5
Step-by-step explanation:
Please answer this correctly
Answer:
42
Step-by-step explanation:
This is a random selection with replacement.
The number of green and purple is equal.
=> Probability to select each: P = 2/4 = 1/2
=> Do the selection 84 times, the best prediction for the number of times picking a purple marble: N = 84 x 1/2 = 42
Hope this helps!
Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean. See Attached Excel for Data. Assume that the population germination time is normally distributed. Find the 97% confidence interval for the mean germination time.
Answer:
97% Confidence interval = (12.62, 18.98)
Step-by-step explanation:
Complete Question
Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean: 18, 12, 20, 17, 14, 15, 13, 11, 21, 17. Assume that the population germination time is normally distributed. Find the 97% confidence interval for the mean germination time.
Solution
We first compute the sample mean and standard deviation for this sample distribution
Sample mean = (Σx)/N = (158/10) = 15.8
Standard deviation = √{[Σ(x - xbar)²]/(N-1)} = 3.3598941782278 = 3.36
Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.
Mathematically,
Confidence Interval = (Sample mean) ± (Margin of error)
Sample Mean = 15.8
Margin of Error is the width of the confidence interval about the mean.
It is given mathematically as,
Margin of Error = (Critical value) × (standard Error of the mean)
Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.
To find the critical value from the t-tables, we first find the degree of freedom and the significance level.
Degree of freedom = df = n - 1 = 10 - 1 = 9
Significance level for 97% confidence interval
(100% - 97%)/2 = 1.5% = 0.015
t (0.015, 9) = 2.9982 (from the t-tables)
Standard error of the mean = σₓ = (σ/√n)
σ = standard deviation of the sample = 3.36
n = sample size = 10
σₓ = (3.36/√10) = 1.0625
97% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]
CI = 15.8 ± (2.9982 × 1.0625)
CI = 15.8 ± 3.1809882246
97% CI = (12.6190117754, 18.9809882246)
97% Confidence interval = (12.62, 18.98)
Hope this Helps!!!
A school librarian purchases a novel for her library. The publisher claims that the book is written at a 5th grade reading level, but the librarian suspects that the reading level is higher than that. The librarian selects a random sample of 40 pages and uses a standard readability test to assess the reading level of each page. The mean reading level of these pages is 5.2 with a standard deviation of 0.8. Do these data give convincing evidence at the = 0.05 significance level that the average reading level of this novel is greater than 5?
Answer:
[tex]t=\frac{5.2-5}{\frac{0.8}{\sqrt{40}}}=1.58[/tex]
The degrees of freedom are given by:
[tex]df=n-1=40-1=39[/tex]
Thep value for this case would be given by:
[tex]p_v =P(t_{(39)}>1.58)=0.061[/tex]
Since the p value is higher than the significance level of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that true mean is not significantly higher than 5.
Step-by-step explanation:
Information provided
[tex]\bar X=5.2[/tex] represent the sample mean
[tex]s=0.8[/tex] represent the sample standard deviation
[tex]n=40[/tex] sample size
[tex]\mu_o =5[/tex] represent the value to verify
[tex]\alpha=0.05[/tex] represent the significance level
t would represent the statistic
[tex]p_v[/tex] represent the p value
Hypothesis to test
We want to verify if the true mean is higher than 5, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 5[/tex]
Alternative hypothesis:[tex]\mu > 5[/tex]
The statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the info given we got:
[tex]t=\frac{5.2-5}{\frac{0.8}{\sqrt{40}}}=1.58[/tex]
The degrees of freedom are given by:
[tex]df=n-1=40-1=39[/tex]
Thep value for this case would be given by:
[tex]p_v =P(t_{(39)}>1.58)=0.061[/tex]
Since the p value is higher than the significance level of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that true mean is not significantly higher than 5.
Alan has reached 25% of his weekly exercise time goal so far this week. If he has exercised for a total of 42 minutes this week, what is his weekly exercise time goal in minutes
Answer:
His weekly exercise time goal is 168 minutes.
Step-by-step explanation:
This question can be solved using a rule of three.
42 minutes is 25% = 0.25 of the total
x minutes is 100% = 1 of the total.
Then
42 minutes - 0.25
x minutes - 1
[tex]0.25x = 42[/tex]
[tex]x = \frac{42}{0.25}[/tex]
[tex]x = 168[/tex]
His weekly exercise time goal is 168 minutes.
Two thousand dollars is deposited into a savings account at 8.5% interest compounded continuously. (a) What is the formula for A(t), the balance after t years? (b) What differential equation is satisfied by A(t), the balance after t years? (c) How much money will be in the account after 5 years? (d) When will the balance reach $3000? (e) How fast is the balance growing when it reaches $3000?
Answer:
a)[tex]A(t)=2000e^{0.085t}[/tex]
b)[tex]A'(t)=170e^{0.085t}[/tex]
c)$3059.1808
d)t=4.77 years
e) The balance growing is $254.99/year
Step-by-step explanation:
We are given that Two thousand dollars is deposited into a savings account at 8.5% interest compounded continuously.
Principal = $2000
Rate of interest = 8.5%
a) What is the formula for A(t), the balance after t years?
Formula [tex]A(t)=Pe^{rt}[/tex]
So,[tex]A(t)=2000e^{0.085t}[/tex]
B)What differential equation is satisfied by A(t), the balance after t years?
So, [tex]A'(t)=2000 \times 0.085 e^{0.085t}[/tex]
[tex]A'(t)=170e^{0.085t}[/tex]
c)How much money will be in the account after 5 years?
Substitute t = 5 in the formula "
[tex]A(t)=2000e^{0.085t}\\A(5)=2000e^{0.085(5)}\\A(5)=3059.1808[/tex]
d)When will the balance reach $3000?
Substitute A(t)=3000
So, [tex]3000=2000e^{0.085t}[/tex]
t=4.77
The balance reach $3000 in 4.77 years
e)How fast is the balance growing when it reaches $3000?
Substitute the value of t = 4.77 in derivative formula :
[tex]A'(t)=170e^{0.085t}\\A'(t)=170e^{0.085 \times 4.77}\\A'(t)=254.99[/tex]
Hence the balance growing is $254.99/year
Which is the solution to the equation 0.5 x + 4.2 = 5.9? Round to the nearest tenth if necessary. 0.9 3.4 5.1 20.2
Answer:
x = 3.4
Step-by-step explanation:
Step 1: Isolate x
0.5x = 1.7
Step 2: Divide both sides by 0.5
x = 3.4
And we have our final answer!
Answer: x=3.4
Step-by-step explanation:
[tex]0.5x+4.2=5.9[/tex]
multiply both sides by 10
[tex]5x+42=59[/tex]
subtract 42 on both sides
[tex]5x=17[/tex]
divide 5 on both sides
[tex]x=\frac{17}{5}[/tex] or
Simplify
x= 3.4
Please answer this question I give brainliest thank you! Number 14
Find the area of a triangle that has the base of 5 inches and a height of 3 3/4 inches
Answer:
9.375 in^2
Step-by-step explanation:
Determine whether the underlined value is a parameter or a statistic. In a national survey of high school students (grades 9 to 12), Modifying 25 % of the students who responded reported that someone had offered, sold, or given them an illegal drug on school property.
Answer:
I am assuming the underlined value is 25%. It is a parameter
Step-by-step explanation:
The value is is a parameter. This is because the parameter is a value that describes the population.
The survey carried out was a national survey of which there were 25% respondents who reported that someone had offered, sold, or given them an illegal drug on school property. It is not a statistics because a sample was not taken out of the population and a survey made on the sample.
The underlined 25% value is the value that summarizes the entire population of high school students
The tensile strength of a certain metal component is normally distributed with a mean of 10,000 kilograms per square centimeter and a standard deviation of 100 kilograms per square centimeter. Measurements are recorded to the nearest 50 kilograms per square centimeter.
Required:
a. What proportion of these components exceed 10,150 kilograms per square centimeter in tensile strength?
b. If specifications require that all components have tensile strength between 9800 and 10,200 kilograms per square centimeter inclusive, what proportion of pieces would we expect to scrap?
Answer:
a. 0.0668
b. 0.9545
Step-by-step explanation:
We have the following information:
mean (m) = 10000
standard deviation (sd) = 100
(a)
We must calculate the proportion of the components exceed 10150 kilograms per square centimeter in tensile strength as follows:
P (x> 10150) = P [(x - m) / sd> (10150 - 1000 /) 100]
P (x> 10150) = P (z> 1.5)
P (x> 10150) = 1 - P (z <1.5)
P (x> 10150) = 1 - 0.9332 (attached table)
P (x> 10150) = 0.0668
Therefore the proportion of the components exceed 10150 kilograms per square centimeter in tensile strength is 0.0668
(b)
We must calculate the proportion of all components has tensile strength between 9800 and 10200, as follows:
P (9800 <x <10200) = P [(9800 - 1000 /) 100 <(x - m) / sd <(10200 - 1000 /) 100]
P (9800 <x <10200) = P (-2 <z <2)
P (9800 <x <10200) = P (z <2) - P (z <-2)
P (9800 <x <10200) = 0.9773 - 0.0228 (attached table)
P (9800 <x <10200) = 0.9545
the proportion of pieces that would expect to scrap is 0.9545
An individual who has automobile insurance from a certain company is randomly selected. Let Y be the number of mov- ing violations for which the individual was cited during the last 3 years. The pmf of Y is y0123 p(y) .60 .25 .10 .05 a. Compute E(Y). b. Suppose an individual with Y violations incurs a sur- charge of $100Y2. Calculate the expected amount of the surcharge
Answer:
a) [tex] E(Y) = \sum_{i=1}^n Y_i P(Y_i)[/tex]
And replacing we got:
[tex] E(Y) =0*0.60 +1*0.25 +2*0.1 +3*0.05 = 0.60[/tex]
b) [tex] E(Y^2) =0^2*0.60 +1^2*0.25 +2^2*0.1 +3^2*0.05 = 1.1[/tex]
And then the expected value would be:
[tex] E(100Y^2) = 100*1.1= 110[/tex]
Step-by-step explanation:
We assume the following distribution given:
Y 0 1 2 3
P(Y) 0.60 0.25 0.10 0.05
Part a
We can find the expected value with this formula:
[tex] E(Y) = \sum_{i=1}^n Y_i P(Y_i)[/tex]
And replacing we got:
[tex] E(Y) =0*0.60 +1*0.25 +2*0.1 +3*0.05 = 0.60[/tex]
Part b
If we want to find the expected value of [tex] 100 Y^2[/tex] we need to find the expected value of Y^2 and we have:
[tex] E(Y^2) = \sum_{i=1}^n Y^2_i P(Y_i)[/tex]
And replacing we got:
[tex] E(Y^2) =0^2*0.60 +1^2*0.25 +2^2*0.1 +3^2*0.05 = 1.1[/tex]
And then the expected value would be:
[tex] E(100Y^2) = 100*1.1= 110[/tex]
A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.
a. What fraction of the calls last between 4.50 and 5.30 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
b. What fraction of the calls last more than 5.30 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
c. What fraction of the calls last between 5.30 and 6.00 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
d. What fraction of the calls last between 4.00 and 6.00 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
e. As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 5 percent of the calls. What is this time? (Round z-score computation to 2 decimal places and your final answer to 2 decimal places.)
Answer:
(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.
(b) The fraction of the calls last more than 5.30 minutes is 0.1271.
(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.
(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.
(e) The time is 5.65 minutes.
Step-by-step explanation:
We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.
Let X = the length of the calls, in minutes.
So, X ~ Normal([tex]\mu=4.5,\sigma^{2} =0.70^{2}[/tex])
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean time = 4.5 minutes
[tex]\sigma[/tex] = standard deviation = 0.7 minutes
(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X [tex]\leq[/tex] 4.50 min)
P(X < 5.30 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{5.30-4.5}{0.7}[/tex] ) = P(Z < 1.14) = 0.8729
P(X [tex]\leq[/tex] 4.50 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{4.5-4.5}{0.7}[/tex] ) = P(Z [tex]\leq[/tex] 0) = 0.50
The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.
Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.
(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)
P(X > 5.30 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{5.30-4.5}{0.7}[/tex] ) = P(Z > 1.14) = 1 - P(Z [tex]\leq[/tex] 1.14)
= 1 - 0.8729 = 0.1271
The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.
(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X [tex]\leq[/tex] 5.30 min)
P(X < 6.00 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{6-4.5}{0.7}[/tex] ) = P(Z < 2.14) = 0.9838
P(X [tex]\leq[/tex] 5.30 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{5.30-4.5}{0.7}[/tex] ) = P(Z [tex]\leq[/tex] 1.14) = 0.8729
The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.
Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.
(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X [tex]\leq[/tex] 4.00 min)
P(X < 6.00 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{6-4.5}{0.7}[/tex] ) = P(Z < 2.14) = 0.9838
P(X [tex]\leq[/tex] 4.00 min) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{4.0-4.5}{0.7}[/tex] ) = P(Z [tex]\leq[/tex] -0.71) = 1 - P(Z < 0.71)
= 1 - 0.7612 = 0.2388
The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.
Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.
(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;
P(X > x) = 0.05 {where x is the required time}
P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{x-4.5}{0.7}[/tex] ) = 0.05
P(Z > [tex]\frac{x-4.5}{0.7}[/tex] ) = 0.05
Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;
[tex]\frac{x-4.5}{0.7}=1.645[/tex]
[tex]{x-4.5}{}=1.645 \times 0.7[/tex]
x = 4.5 + 1.15 = 5.65 minutes.
SO, the time is 5.65 minutes.
What statistical test and how many of them will need to be used to explore the difference in average cholesterol within each group before and after the treatment (i.e., how much did cholesterol level change within each group as result of the treatment)?
Answer:
The statistical test to be used is the paired t-test.
Step-by-step explanation:
The dependent t-test (also known as the paired t-test or paired-samples t-test) compares the two means associated groups to conclude if there is a statistically significant difference between these two means.
We use the paired t-test if we have two measurements on the same item, person or thing. We should also use this test if we have two items that are being measured with a unique condition.
For instance, an experimenter tests the effect of a medicine on a group of patients before and after giving the doses.
Similarly, in this case a paired t-test would be used to deter whether there was any changes in the cholesterol level within each group as result of the treatment.
Thus, the statistical test to be used is the paired t-test.
Number of multiples of 7 between 200 and 1000
Answer:
114
Step-by-step explanation:
Answer:
144Step-by-step explanation:
The Giant Machinery has the current capital structure of 65% equity and 35% debt. Its net income in the current year is $250,000. The company is planning to launch a project that will requires an investment of $175,000 next year. Currently the share of Giant machinery is $25/share. Required: a) How much dividend Giant Machinery can pay its shareholders this year and what is dividend payout ratio of the company? Assume the Residual Dividend Payout Policy applies. (4 marks) b) If the company is paying a dividend of $2.50/share and tomorrow the stock will go ex-dividend. Calculate the ex-dividend price tomorrow morning. Assuming the tax on dividend is 15%. (2 marks) c) Little Equipment for Hire is a subsidiary in the Giant Machinery and currently under the liquidation plan due to the severe contraction of operation due to corona virus. The company plans to pay total dividend of $2.5 million now and $ 7.5 million one year from now as a liquidating dividend. The required rate of return for shareholders is 12%. Calculate the current value of the firm’s equity in total and per share if the firm has 1.5 million shares outstanding. (4 marks
Answer:
The amount of dividend Giant Machinery can pay its shareholders this year = $136,250
The Dividend Payout ratio of the company = 54.50%
Ex-dividend price = $22.875
The current value of the firm's equity in total = $9196428.571
The current value of the firm's equity per share = $6.131
Step-by-step explanation:
Given that:
The Current Capital Structure = 65% equity and 35% debt.
The net income in the current year = $250,000
Also, the company is planning to launch a project that will requires an investment of $175,000 next year.
The current share price = $25/share
(a)
The first objective is to determine how much dividend Giant Machinery can pay its shareholders this year and what is dividend payout ratio of the company.
The amount of dividend Giant Machinery can pay its shareholders this year= Net profit - Investment amount × (percentage of equity)
= 250,000 - 175,000 × (65%)
= 250,000 - 113,750
= $136,250
The dividend payout ratio of the company.can be calculated as follows:
Dividend Payout ratio of the company [tex]= \mathbf{\dfrac{ total \ \ dividends}{total \ \ earning}}[/tex]
We all know that the net income in the current year is the Total Earning which is 250,000
Thus;
The Dividend Payout ratio of the company [tex]= \mathbf{\dfrac{ 136250}{250000}}[/tex]
The Dividend Payout ratio of the company = 0.545
The Dividend Payout ratio of the company = 54.50%
b).
If the company is paying a dividend of $2.50/share and tomorrow the stock will go ex-dividend.
Also, assuming:
the tax on dividend = 15%.
We are to calculate the ex-dividend price tomorrow morning.
To calculate the ex-dividend price tomorrow morning; we use the relation:
Ex-dividend price which is the result of the difference between the current price and dividend multiply by the difference between the 1 and the tax on the dividend
i.e
Ex-dividend pric = current price - Dividend × (1 - tax on dividend)
Given that :
The current share price = $25/share
Therefore;
Ex-dividend price = $25 - $2.5 × ( 1 - 15%)
Ex-dividend price = $25 - $2.5 × ( 1 - 0.15)
Ex-dividend price = $25 - $2.5 × 0.85
Ex-dividend price = $25 - $2.125
Ex-dividend price = $22.875
C)
From the information given in the part C of the question:
the company now plans to pays a total dividend of $2.5 million.= $ 2,500.000
Let say the year the dividend was paid was 0 year
and 7.5 million one year from now as a liquidating dividend.
So; the dividend paid in year 1 now = 7.5 million = $ 7,500,000
The required rate of return for shareholders is 12%
Outstanding share of the firm = 1.5 million = $1,500,000
We are to calculate the current value of the firm’s equity in total and per share if the firm has 1.5 million shares outstanding.
To start with the current value of the firm’s equity in total;
The current value of the firm's equity in total =[tex]\mathbf{amount \ of \ dividend \ paid \ in \ 0 \ year + \dfrac {amount \ of \ dividend \ paid \ in \ 1 \ year } { 1+required \ rate \ of \ return } }[/tex]= [tex]\mathbf{2,500,000 + \dfrac{7,500,000 }{1+ 0.12} }[/tex]
[tex]=\mathbf{2,500,000 + \dfrac{7,500,000 }{1.12} }[/tex]
= 2,500,000 + 6696428.571
= $ 9196428.571
The current value of the firm's equity per share =[tex]\mathbf{ \dfrac{ current \ value \ of \ the \ firm's \ equity \ in \ total}{ Outstanding \ share }}[/tex]
[tex]\mathbf{= \dfrac{9196428.571}{1500000}}[/tex]
= $6.130952381
≅ $6.131
The current value of the firm's equity per share = $6.131
Someone flips five coins, but you don’t see the outcome. The person reports that no tails are showing. What is the probability that the person flipped 5 heads?
Answer:
0.03125 = 3.125% probability that the person flipped 5 heads
Step-by-step explanation:
For each coin, there are only two possible outcomes. Either it was heads, or it was tails. The result of a coin toss is independent of other coin tosses. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Five coins:
This means that n = 5.
Fair coin:
Equally as likely to be heads or tails, so p = 0.5.
What is the probability that the person flipped 5 heads?
This is P(X = 5).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125[/tex]
0.03125 = 3.125% probability that the person flipped 5 heads
The mean number of hours of part-time work per week for a sample of 317 teenagers is 29. If the margin of error for the population mean with a 95% confidence interval is 2.1, construct a 95% confidence interval for the mean number of hours of part-time work per week for all teenagers.
Answer:
The degrees of freedom are given by:
[tex]df=n-1=317-1=316[/tex]
And replaicing we got:
[tex]29-2.1=26.9[/tex]
[tex]29+2.1=31.1[/tex]
The 95% confidence interval would be between 26.9 and 31.1
Step-by-step explanation:
Information given
[tex]\bar X= 29[/tex] represent the sample mean
[tex]\mu[/tex] population mean
s represent the sample standard deviation
[tex] ME= 2.1[/tex] represent the margin of error
n represent the sample size
Solution
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
And this formula is equivalent to:
[tex] \bar X \pm ME[/te]x
The degrees of freedom are given by:
[tex]df=n-1=317-1=316[/tex]
And replaicing we got:
[tex]29-2.1=26.9[/tex]
[tex]29+2.1=31.1[/tex]
The 95% confidence interval would be between 26.9 and 31.1
Find the area of the smaller sector.
A
6 in
030°
Area = [? ]in?
B
Round your answer to the nearest hundredth.
Answer:
9.42 in²
Step-by-step explanation:
The area of whole circle S=pi*R² , where pi is appr. 3.14, R= 6 in
S= 3.14*6² =113.04 in²
The area of smaller sector is Ssec=S/360*30=113,04/12=9.42 in²
The area of the smaller sector with a central angle of 30 degrees and a radius of 6 inches is 9.42478 square inches.
To find the area of a sector, you can use the formula:
Area of sector = (θ/360) × π × r²
where θ is the central angle in degrees, r is the radius of the sector.
The central angle is 30 degrees and the radius is 6 inches.
Plugging these values into the formula:
Area of sector = (30/360) × π × 6²
= (1/12) × π × 36
= (1/12) × 3.14159 × 36
= 9.42478 square inches
To learn more on Area of sector click:
https://brainly.com/question/29055300
#SPJ2
The annual interest on a $14,000 investment exceeds the interest earned on a $7000 investment by $595 . The $14,000 is invested at a 0.5% higher rate of interest than the $7000 . What is the interest rate of each investment?
Answer:
x= 8.00 Interest rate on $14000
y= 7.50 Interest rate on $7000
Step-by-step explanation:
Let interest rate of $14000 be x%
and Interest rate for $7000 be y %
According to the first condition
14000 * x% - 7000 * y% = 595
multiply by 100
14000x-7000y = 59500
/700
20x-10y=85.................(1)
II condition
x%=y%+0.5%
x=y+0.5
x-y=0.5..................................(2)
solve (1) & (2)
20 x -10 y = 85 .............1
Total value
1 x -1 y = 0.50 .............2
Eliminate y
multiply (1)by 1
Multiply (2) by -10
20.00 x -10.00 y = 85.00
-10.00 x + 10.00 y = -5.00
Add the two equations
10.00 x = 80.00
/ 10.00
x = 8.00
plug value of x in (1)
20.00 x -10.00 y = 85.00
160.00 -10.00 y = 85.00
-10.00 y = 85.00 -160.00
-10.00 y = -75.00
y = 7.50
x= 8.00 Interest rate on $14000
y= 7.50 Interest rate on $7000
Judy wants to buy a new bed, but it will cost her $668. The sales tax rate is 4.4% where she is buying the bed. Her budget only allows her to purchase a bed that is no more than $700. the sales tax for her. Round to two decimal places.
Answer:
$41.21
Step-by-step explanation: